Source: MTH 132, Michigan State University, Section 2.1
Tags: derivative, limit definition, tangent line, differentiable, non-differentiable, corner, cusp, vertical tangent, MTH 132, Calculus I, MSU
Difficulty: Intermediate | Prerequisites: Limits (Sec. 1.4 -- 1.6), continuity (Sec. 1.8), basic algebra and rationalisation techniques
The derivative is the central concept of differential calculus. It measures the instantaneous rate of change of a function, and its geometric meaning is the slope of the tangent line. In MTH 132 you are expected to compute derivatives from the limit definition before learning shortcut rules, and to recognise situations where a derivative fails to exist. This topic builds directly on your limit skills and connects forward to every differentiation technique you will learn.
The derivative f'(a) is defined as the limit of [f(a+h) - f(a)] / h as h approaches 0. Geometrically it gives the slope of the tangent line at x = a. A function can fail to be differentiable at a point if it has a corner (like |x|), a cusp, a vertical tangent, or a discontinuity.
Derivative at a point, f'(a)
The limit f'(a) = lim(h->0) [f(a+h) - f(a)] / h, provided the limit exists. In simple terms, it is the slope of the function at exactly that one point.
Tangent line
The line that just touches the curve at a single point, with slope equal to the derivative there. Its equation is y - f(a) = f'(a)(x - a).
Differentiable at a point
A function is differentiable at x = a if f'(a) exists (the limit definition gives a finite number). Think of it as: the graph is smooth enough at that point to have a well-defined slope.
Non-differentiable
A function is not differentiable at x = a when the limit definition fails. This happens at corners (left and right slopes disagree), cusps, vertical tangent lines (slope is infinite), or discontinuities.
Rationalisation (conjugate multiplication)
An algebraic technique used when the limit definition produces a difference of square roots. Multiply numerator and denominator by the conjugate to clear the radical. Essential for computing derivatives of functions like sqrt(x) from the definition.
Write out the definition: f'(a) = lim(h->0) [f(a+h) - f(a)] / h.
Substitute f(a+h) and f(a), expand, simplify, cancel the h in the denominator, then take the limit as h -> 0.
Worked example: f(x) = 3x + 4, find f'(4).
f(4+h) = 3(4+h) + 4 = 16 + 3h.
f(4) = 16.
[f(4+h) - f(4)] / h = 3h / h = 3.
lim(h->0) 3 = 3. So f'(4) = 3.
f(x) = sqrt(x). Use the definition to find f'(1).
f'(1) = lim(h->0) [sqrt(1+h) - sqrt(1)] / h.
Multiply numerator and denominator by the conjugate: (sqrt(1+h) + 1).
Numerator becomes (1+h) - 1 = h. Denominator becomes h(sqrt(1+h) + 1).
Cancel h: 1 / (sqrt(1+h) + 1). As h -> 0, this gives 1/2.
Tangent line: y - 1 = (1/2)(x - 1), or equivalently y = (1/2)x + 1/2.
f(x) = 3/x. Find f'(3) from the definition.
f'(3) = lim(h->0) [3/(3+h) - 3/3] / h = lim(h->0) [3/(3+h) - 1] / h.
Combine the fraction in the numerator: [3 - (3+h)] / [h(3+h)] = -h / [h(3+h)].
Cancel h: -1/(3+h). As h -> 0, this gives -1/3.
Tangent line: y - 1 = (-1/3)(x - 3), or equivalently y = (-1/3)x + 10/3.
Corner (sharp turn): f(x) = |x - 4| at x = 4.
The left-hand derivative limit gives -1.
The right-hand derivative limit gives +1.
Since -1 ≠ 1, f'(4) does not exist.
Vertical tangent: g(x) = (x - 4)^(1/3) at x = 4.
The derivative limit tends to infinity (the tangent line is vertical).
A vertical line has undefined slope, so g'(4) does not exist.
Other reasons for non-differentiability include cusps and discontinuities, though the recitation focused on corners and vertical tangents.
f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}Tangent line through (a, f(a)) with slope m = f'(a):
y - f(a) = f'(a)(x - a)Conjugate trick for square roots: multiply top and bottom by (sqrt(a+h) + sqrt(a)) to clear the radical difference in the numerator.
The derivative as a rate of change is everywhere: velocity is the derivative of position, current is the derivative of charge, and marginal cost is the derivative of total cost. When you compute f'(a) from the limit definition, you are doing exactly what a physicist does when measuring instantaneous velocity from a position function.
Students often skip the limit step and just "plug in h = 0" from the start. You cannot do that because it gives 0/0. The whole point is to simplify algebraically until the h in the denominator cancels, and only then take the limit.
Students confuse the derivative (a number, the slope) with the tangent line (an equation of a line). f'(a) is the slope; the tangent line uses that slope in the point-slope formula.
Students assume that if a function is continuous, it must be differentiable. Continuity is necessary but not sufficient. f(x) = |x| is continuous everywhere but not differentiable at x = 0.
When rationalising, students sometimes forget to multiply the denominator as well. Both top and bottom must be multiplied by the conjugate.
⚠️ "Use the limit definition to find f'(a)" is a guaranteed exam question in MTH 132. You must show the full limit process, not use shortcut rules.
⚠️ Finding the equation of the tangent line from the definition is a two-part problem: compute f'(a), then write y - f(a) = f'(a)(x - a). Do not forget to state the final equation.
⚠️ Graphing two functions that fail to be differentiable at the same point for different reasons (corner vs. vertical tangent) is a common exam exercise. Be able to sketch both and explain the difference.
Fill in the blank: f'(a) = lim(h->0) ______.
[f(a+h) - f(a)] / h.
True or False: If f is differentiable at x = a, then f is continuous at x = a.
True.
True or False: If f is continuous at x = a, then f is differentiable at x = a.
False. Counterexample: f(x) = |x| at x = 0.
Fill in the blank: The tangent line to y = f(x) at x = a has slope ______ and passes through the point ______.
Slope f'(a); point (a, f(a)).
Q: Use the limit definition to find f'(4) for f(x) = 3x + 4.
A: f'(4) = lim(h->0) [3(4+h)+4 - (3(4)+4)] / h = lim(h->0) 3h/h = 3.
Q: Find the equation of the tangent line to y = sqrt(x) at (1, 1) using the limit definition.
A: f'(1) = 1/2 (via rationalisation). Tangent line: y - 1 = (1/2)(x - 1), i.e. y = (1/2)x + 1/2.
Q: Find the equation of the tangent line to y = 3/x at (3, 1) using the limit definition.
A: f'(3) = -1/3. Tangent line: y - 1 = (-1/3)(x - 3), i.e. y = (-1/3)x + 10/3.
Q: Give two different reasons a function might not be differentiable at x = 4, with an example of each.
A: (1) Corner: f(x) = |x - 4|. Left-hand derivative is -1, right-hand is +1, so f'(4) DNE. (2) Vertical tangent: g(x) = (x - 4)^(1/3). The derivative limit tends to infinity, so g'(4) DNE.
Q: What algebraic technique is used to compute the derivative of sqrt(x) from the definition?
A: Multiply numerator and denominator by the conjugate, sqrt(1+h) + sqrt(1), to eliminate the radical difference.
The limit definition of the derivative is the theoretical foundation for every differentiation rule you will learn next (power rule, product rule, chain rule). Those rules are shortcuts derived from this limit, so understanding the definition helps you see why the rules work.
Differentiability also connects back to continuity: differentiable at a point implies continuous at that point, but not the reverse. This relationship is tested frequently, and the standard counterexample is f(x) = |x| at x = 0.
derivative, limit definition of derivative, difference quotient, tangent line, point-slope form, slope at a point, differentiable, non-differentiable, corner, cusp, vertical tangent line, rationalisation, conjugate multiplication, f'(a), instantaneous rate of change, MTH 132, Calculus I, MSU, Section 2.1